\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
 2006 International Conference in Honor of Jacqueline Fleckinger.
\newline {\em Electronic Journal of Differential Equations},
Conference 16, 2007, pp. 95--102.
\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or
http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2007 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document} \setcounter{page}{95}

\title[\hfilneg EJDE/Conf/16 \hfil
Asymptotic behavior of a dry friction oscillator]
{Asymptotic behavior of a weakly forced dry friction oscillator}

\thanks{J.~I. D\'{\i}az was supported by
         project MTM2005-03463 from DGISGPI, Spain}

\author[J. Ildefonso D\'{\i}az,  G. Hetzer
\hfil EJDE/Conf/16 \hfilneg]
{J. Ildefonso D\'{\i}az, Georg Hetzer}

\address{J. Ildefonso D\'{\i}az \newline
Departamento de Matem\'{a}tica Aplicada,
Universidad Complutense de Madrid,
28040 Madrid, Spain}
\email{ji\_diaz@mat.ucm.es}

\address{Georg Hetzer \newline
Department of Mathematics and Statistics,
Auburn University, AL 36849-5310, USA}
\email{hetzege@auburn.edu}

\thanks{Published May 15, 2007.}
\subjclass[2000]{34A60, 34D05}
\keywords{Nonlinear oscillators and systems; dry friction; dead zone;
 \hfill\break\indent     extinction in finite time; stick-slip motion}

\dedicatory{Dedicated to  Jacqueline Fleckinger  on the occasion of\\
 an international conference in her honor}


\begin{abstract}
 This note is devoted to stick-slip aspects of the
 motion of a dry friction damped oscillator under weak irregular
 forcing. Our main result complements \cite[Theorem 3.(a)]{DHS} and
 is also related to \cite{AmDi}, where a non-Lipschitz model for
 Coulomb friction was consider in the unforced case. We provide
 sufficient conditions guaranteeing that solutions stabilizing in
 finite time, but observe also an infinite succession of
 ``stick-slip'' behavior. The last section discusses an extension to
 certain systems of such oscillators.
\end{abstract}


\maketitle

\newtheorem{theorem}{Theorem}[section]
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}

\newcommand\abs[1]{|#1|}
\newcommand\norm[1]{\|#1\|}

\section{Introduction}

Its is well known that the abstract Cauchy problem associated with
multivalued monotone (resp.~accretive) operators on Hilbert spaces
(resp.~Banach spaces) may lead to very peculiar strong convergence
asymptotic behaviour for its solutions. More precisely, if for
instance $X=H$ is a Hilbert space, and
$A:D(A)\to\mathcal{P}(H)$ is a maximal monotone operator
multivalued at $0$ (with $0\in intD(A)$) then the solution of the
\begin{equation}
\begin{gathered}
\frac{du}{dt}(t)+Au(t)\ni f(t) \quad \text{in }X,\\
u(0)=u_{0},
\end{gathered}
\end{equation}
possesses the property of extinction in finite time once
we assume that $f$ satisfies
\begin{equation}
B(f(t),\epsilon)\subset A0,\quad
\text{for a.e. $t\geq t_{f}$, for some $\epsilon>0$ and
$t_{f}\geq 0$}. \label{multivalued}
\end{equation}
This result, due to H. Brezis (\cite{Brezis-Montreal}), has been
generalized in many different directions in the last twenty years
(see, for instance, the survey \cite{Diaz}). The main goal of this
paper is to investigate some simple cases in which the image of $0$
under the multivalued operator is not so large as to contain a ball.

In order to get some insight into this type of difficulties we shall
first study the long-term behaviour of solutions of
\begin{equation}\label{DE}
\ddot{x}+x-p(t)\in\mathop{\rm sgn}(-\dot{x}),
\end{equation}
where
$$
\mathop{\rm sgn}(y):=\begin{cases}
 y/|y| & \text{if } y\ne 0 \\
{[-1,1]} & \text{if }y=0
\end{cases}
$$
represents the damping force due to dry friction and
$p\in C([0,\infty),\mathbb{R})$ is an external
forcing, which is weak in the sense that
$\sup\{\abs{p(t)}:t\in[0,\infty)\}<1$. \eqref{DE} is one of a
variety of damped oscillator equations modelling dry friction, and
our interest in this particular setting arises from the fact that it
describes pure dry friction damping, mathematically, the more
challenging case. We refer to \cite{AmDi}, \cite{Bo} and \cite{Ku}
for other settings and references, to \cite{DHS} and \cite{Ku} for
``resonance'' under almost periodic forcing, and to \cite{CDB} and
the references therein for dry friction damped wave equations. As
for our purposes, it was shown in \cite{DHS} (formally in an almost
periodic setting), that every solution converges to a constant
solution as $t\to\infty$ (cf.~also Lemma \ref{L1} (4) below). We are
interested in solutions which either are eventually constant (the
mass comes to rest in finite time) or show an infinite succession of
stick-slip events. We allow temporally irregular forcing and can
require without loss of generality that
$$
\overline{p}:=\limsup_{t\to\infty} p(t)=-\underline{p}
:=\liminf_{t\to\infty} p(t).
$$

\begin{definition}\label{DZ} \rm
Let $a,\ b\in\mathbb{R}_+$, $a<b$. An interval $[a,b]$
$\bigl([a,\infty)\bigr)$ is called a dead zone of a solution $x$ of
\eqref{DE}, if $\dot{x}(t)=0$ for $t\in [a,b]$
$\bigl(t\in[a,\infty)\bigr)$.
\end{definition}

 Our main result regarding \eqref{DE} read as follows.

\begin{theorem}\label{MT}
If $x$ is a solution of \eqref{DE} with
$x_\infty:=\lim_{t\to\infty} x(t)<1+\overline{p}$. Then one
of the following alternatives occurs.
\begin{enumerate}
 \item
$\overline{p}-1<x_\infty<1-\overline{p}$, $t\mapsto x_\infty$ is a
constant solution of \eqref{DE}, and $x$ has a dead zone
$[\underline t,\infty)$.

\item $\abs{x_\infty}>1-\overline{p}$, $x$ is monotone
and has a compact dead zone in each neighborhood of infinity.

\item If $\abs{x_\infty}=1-\overline{p}$, there may or may not
be a dead zone.
\end{enumerate}
\end{theorem}

We conclude this paper with some partial result concerning the
system
\begin{equation} \label{PN}
\begin{gathered}
m\ddot{x}_{i}(t)+k(-x_{i-1}(t)+2x_{i}(t)-x_{i+1}(t))+\mu_{\beta}\mathop{\rm
sgn}(\dot
{x}_{i}(t))\ni p_{i}(t)\\
x_{i}(0)=u_{0,i},\mbox{ }\dot{x}_{i}(0)=v_{0,i}
\end{gathered}
\end{equation}
$i=1,\dots,N,$ where we are assuming that
$$
x_{0}(t)=0,\mbox{ }x_{N+1}(t)=1\quad\mbox{for  }t\in(0,+\infty),
$$
$m,\mu_{\beta}$ are positive constants and the term
$\mu_{\beta}\mathop{\rm sgn}(\dot {x}_{i}(t))$ represents the
Coulomb friction. This system arises in the modeling of the
vibration of $N$-particles of equal mass $m$ in a non-inertial
coordinate system. Indeed, we denote the locations of the particles,
along the interval $(0,1)$ of the $x$ axis, by $x_{i}(t)$, and we
assume that each particle is connected to its neighbors by two
harmonic springs of strength $k$. We also assume that the particles
are subject to a resultant friction force (the Coulomb (or solid)
friction). Functions $p_{i}(t)$ correspond to fictitious forces due
to the change of variable with respect to an inertial system. We
show that, at least, in some special cases the first conclusion of
Theorem \ref{MT} remains true.

\section{Preliminaries} It is worth noting that the formally more general
equation
\begin{equation}\label{DE2}
\ddot{x}+kx-p(t)\in\mu\mathop{\rm sgn}(-\dot{x})
\end{equation}
with $k$, $\mu\in(0,\infty)$, and $p\in
L^\infty([0,\infty),\mathbb{R})$ can be re-scaled to the simpler form
\eqref{DE}. Replacing  $x$ by ${x\over\mu}$ and $p$ by ${p\over\mu}$
yields $\mu=1$ without loss of generality. Next, let
$\bar{p}:=\limsup_{t\to\infty} p(t)$ and
$\underline{p}:=\liminf_{t\to\infty} p(t)$. The
transformation $x\to x-{{\bar{p}+\underline{p}}\over{2k}}$ and $p\to
p-{{\bar{p}+\underline{p}}\over{2}}$ allows us to assume
$\underline{p}=-\bar{p}$. Finally, setting $\tau={t\over{\sqrt{k}}}$
and $y(\tau)=k x({t\over{\sqrt{k}}})$, one obtains
$\dot{y}(\tau)=\sqrt{k} \dot{x}({t\over{\sqrt{k}}})$ and
$\ddot{y}(\tau)=\ddot{x}({t\over{\sqrt{k}}})$, hence
$\ddot{y}+y-p(\tau)\in-\mathop{\rm sgn}(-\dot{y})$, thus we arrive at \eqref{DE}
under the additional ``symmetry hypotheses''
$\bar{p}=-\underline{p}$.

For the rest of this article, we consider \eqref{DE} under the Hypotheses:
$$
p\in L^\infty(\mathbb{R}_+,\mathbb{R})\cap C(\mathbb{R}_+,\mathbb{R}), \quad
\overline{p}=-\underline{p}.
$$

\begin{definition}\label{Sol} \rm
One calls $x\in W^{2,1}_{\rm loc}([0,\infty),\mathbb{R})$ a
solution of \eqref{DE}, if there exists a $u\in L^\infty(\mathbb{R}_+,\mathbb{R})$
with $u(t)\in \mathop{\rm sgn}(\dot{x}(t))$ for a.e.~$t\in\mathbb{R}_+$ such that
$\ddot{x}(t)+x(t)=p(t)-u(t)$ for a.e.~$t\in(0,\infty)$.
\end{definition}

 Note that $\abs{u(t)}\le 1$ for a.e. $t\in(0,\infty)$.
The general theory of ``multi-valued'' ordinary
differential equations (\cite[\S 5]{De}, \cite[section 2.2]{Ku})
yields the following result.

\begin{proposition}\label{EU}
The initial-value problem \eqref{DE}, $x(t_0)=x_0$, $\dot x(t_0)=\eta_0$ has for each
$(t_0,x_0,\eta_0)\in\mathbb{R}^3$ a (forwardly) unique ``global'' solution
$x\in W^{2,1}_{\rm loc}([t_0,\infty),\mathbb{R})$.
\end{proposition}

 As mentioned, we are interested in whether solutions
develop dead zones. The first example shows that ``stronger''
forcing limits the length of dead zones.

\begin{example} \rm
Let $p(t)=(1+{1\over{1+t}})\sin(t)$. Then $\bar{p}=1$, but a
solution of \eqref{DE} cannot have dead zones of length greater than
$\pi$, since $\abs{p({\pi\over2}+j\pi)-p({\pi\over2}+(j-1)\pi)}>2$
for $j\in\mathbb{N}$.
\end{example}

 The next example indicates that the second alternative of
Theorem \ref{MT} can in fact occur.

\begin{example} \rm
Let $\overline{p}\in (0,1)$, $x_0\in
(1-\overline{p},1+\overline{p})$, $0<t_1<t_2<t_3$, and
\[
p(t)=\begin{cases}\overline{p}& t\in [0,t_1),\\
-\overline{p} & t\in[t_1,t_2),\\ \overline{p} &t\in[t_2,t_3].
\end{cases}
\]
If $t_2-t_1<{\pi\over 4}$ is sufficiently small, then
$2\overline{p}-(x_0-1+ \overline{p})\cos(t_2-t_1)>0$ in view of
$x_0<1+\overline{p}$. Moreover, let
$$
t^*_2=t_2+\arctan({{(x_0-1
+\overline{p})\sin(t_2-t_1)}\over{2\overline{p}-(x_0-1+
\overline{p})\cos(t_2-t_1)}})
$$
satisfy $t_2<t_2^*<t_3$. One verifies that
\[
x(t):=\begin{cases} x_0&t\in[0,t_1],\\
(x_0-1+ \overline{p})\cos(t-t_1)+1-\overline{p}&t\in(t_1,t_2),\\
\Bigl((x_0-1+ \overline{p})\cos(t_2-t_1)- 2\overline{p}\Bigr)
\cos(t-t_2) \\
+1-P-(x_0-1+ \overline{p})\sin(t_2-t_1)\sin(t-t_2)
 &t\in(t_2,t_2^*).
\end{cases}
\]
 solves \eqref{DE} on $[0,t_2^*)$ and satisfies $x(0)=x_0$,
$\dot x(0)=0$. In fact, one has
$\dot x(t)=-(x_0-1+\overline{p})\sin(t-t_1)<0$ for $t\in [t_1,t_2]$, $\dot
x(t)=-\Bigl((x_0-1+\overline{p})\cos(t_2-t_1)- 2\overline{p}\Bigr)
\sin(t-t_2)-(x_0-1+ \overline{p})\sin(t_2-t_1)\cos(t-t_2)<0$ for
$t\in(t_2,t_2^*)$, and $\dot x(t_2^*)=0$.

 Finally, $\bigl((x_0-1+ \overline{p})\cos(t_2-t_1)-
2\overline{p}\bigr) \cos(t_2^*-t_2)  +1-P-(x_0-1+
\overline{p})\sin(t_2-t_1)\sin(t_2^*-t_2)\to x_0$ as $t_2\to t_1$.
Therefore, given $0<\epsilon<x_0-1+\overline{p}$, we can choose
$t_2$ such that $x_0-x(t_2^*)<\epsilon/2$ and have $x(t)=x(t_2^*)$
for $t\in (t_2^*,t_3]$. We can repeat this process with
${\epsilon\over {2^j}}$ in the $j-th$ step and obtain a solution of
\eqref{DE} on $[0,\infty)$ with a $p$ that switches between $\pm 1$.
The solution converges to an $x_\infty >1-\overline{p}$, is not
eventually constant, and has infinitely many dead zones near
infinity.
\end{example}

It is easy to see how to modify the example in order to find
stick-slip behavior for a smooth $p$. One has to guarantee that $p$
takes on the value $-1$ in each of the intervals
$(t_{2j-1},t_{2j})$, which prevents infinitely long dead zones, but
that these intervals are so short that
$x(t_{2j-1})-x(t_{2j+1})<\frac{\epsilon}{2^j}$ for $j\in\mathbb{N}$.

 Next we collect some folklore results (cf.~also \cite{DHS} or
\cite{Ku}), which we prove for the reader's convenience.

\begin{lemma}\label{L1}
Let $\norm{p}_\infty<1$ and $x$ and $y$ be solutions of \eqref{DE}.
Then
\begin{enumerate}
\item $t\mapsto \dot{x}(t)^2+x(t)^2$ is nonincreasing on $\mathbb{R}_+$
and strictly decreasing on intervals which do not intersect the
interior of any dead zone of $x$.
\item $\int_0^\infty\abs{\dot{x}(t)}\ dt\le {1\over{2(1-\norm{p}_\infty})}
\Bigl[\dot{x}(0)^2+x(0)^2\Bigr]$.
\item
$\norm{\ddot{x}}_\infty\le
1+\norm{p}_\infty+\sqrt{\dot{x}(0)^2+x(0)^2}$.
\item $x_\infty:=\lim_{t\to\infty} x(t)$ exists and
belongs to $-1-\norm{p}_\infty\le x_\infty\le 1+\norm{p}_\infty$.
Moreover, $\dot{x}(t)\to 0$ as $t\to\infty$.
\item $(\dot{x}-\dot{y})^2+(x-y)^2$ is nonincreasing on
$\mathbb{R}_+$.
\item The interval $[\sup p-1,\inf p+1]$
forms the set of constant solutions of \eqref{DE}.
\end{enumerate}
\end{lemma}

\begin{proof}
Let $u\in L^\infty(\mathbb{R}_+,\mathbb{R})$ with $u\in\mathop{\rm sgn}\circ\dot{x}$ a.e. One
has
\begin{equation}\label{EE}
{1\over2}{d\over{dt}}\Bigl[\dot{x}(t)^2+x(t)^2\Bigr]=\dot{x}(t)
\Bigl[p(t)-u(t)\Bigr]\le-\abs{\dot{x}(t)}+\abs{p(t)}\abs{\dot{x}(t)}
\le-\abs{\dot{x}(t)}\bigl(1-\norm{p}_\infty\bigr).
\end{equation}
 This yields the first assertion of 1. Also, if
$0<t_1<t_2<\infty$ with
$\dot{x}(t_1)^2+x(t_1)^2=\dot{x}(t_2)^2+x(t_2)^2$, then
$-\abs{\dot{x}(t)}+p(t)\dot{x}(t)=0$ for $t\in [t_1,t_2]$, hence
$\dot{x}(t)=0$ for $t\in [t_1,t_2]$ in view of $\norm{p}_\infty<1$.

2. One obtains from \eqref{EE} that $\int_0^t
\abs{\dot{x}(s)}\ ds\le {1\over{2(1-\norm{p}_\infty})}
\bigl[\dot{x}(0)^2+x(0)^2\bigr]$.

3. Inequality \eqref{EE} implies $\norm{x}_\infty^2\le \dot{x}(0)^2+x(0)^2$,
hence \eqref{DE} yields the $L^\infty$-bound for $\ddot x$.

4. Statement 2 and $\abs{x(\bar{t})-x(\underline{t})}\le
\int_{\underline{t}}^{\bar{t}} \abs{\dot{x}}\ dt$ for
$0\le\underline{t}<\bar{t}<\infty$ imply the convergence of $x(t)$
as $t\to\infty$. Since $\dot{x}(t)^2+x(t)^2$ also converges as
$t\to\infty$, $\abs{\dot{x}(t)}$ converges, and its limit is equal
to 0 in view of $\dot{x}\in L^1(\mathbb{R}_+,\mathbb{R})$. Finally, let $u\in
\mathop{\rm sgn}(\dot{x})$ such that $\ddot{x}(t)+x(t)=p(t)-u(t)$ for
a.e.~$t\in\mathbb{R}_+$. Assume that $x_\infty>1+\norm{p}_\infty$. Select
$\epsilon\in(0,1-\norm{p}_\infty)$ with
$x_\infty>1+\norm{p}_\infty+2\epsilon$ and choose $\underline{t}\ge
0$ such that $x(t)\ge x_\infty-\epsilon$ and
$\abs{\dot{x}}<\epsilon$ for $t\in[\underline{t},\infty)$. Since
$p(t)-x(t)-u(t)\le p(t)-x_\infty+\epsilon+1\le
p(t)-(1+\norm{p}_\infty+2\epsilon)+\epsilon+1 \le -\epsilon$ for
a.e.~$t\in[\underline{t},\infty)$, one has $\ddot{x}<-\epsilon$
a.e.~on $[\underline{t},\infty)$, hence $\dot{x}\le
-\epsilon(t-\underline{t})$ a.e.~on $[\underline{t},\infty)$  which
is a contradiction. Likewise, one obtains $x_\infty\ge
-1-\norm{p}_\infty$.

5. Let $v\in L^\infty(\mathbb{R}_+,\mathbb{R})$ with $v\in\mathop{\rm sgn}\circ\dot{y}$, then
\begin{equation}\label{EE2}
{1\over2}{d\over{dt}}\Bigl[(\dot{x}(t)-\dot{y}(t))^2+(x(t)-y(t))^2\Bigr]
=-\bigl(u(t)-v(t)\bigr)\bigl(\dot{x}(t)-\dot{y}(t)\bigr)\le 0
\end{equation}
 for $t\in (0,\infty)$ a.e.

6. If $z\in [\sup p -1,\inf p+1]$, then $-1\le z-p(t)\le 1$, hence
$z-p(t)\in\mathop{\rm sgn}(0)$ for all $t\in (0,\infty)$, which shows that $z$
is a constant solution of \eqref{DE}. On the other hand, if $z$ is a
constant solution of \eqref{DE}, then $z-p(t)\in\mathop{\rm sgn}(0)$ for all
$t\in (0,\infty)$, hence $-1\le z-p(t)\le 1$  for all $t\in
(0,\infty)$, thus, $-1\le z-\sup p$ and $z-\inf p\le 1$.
\end{proof}

 As for statement 6, the following example shows that one
can have solutions with dead zones of the form $[a,\infty)$ which
stay away from the set of constant solutions. Obviously, the reason
is that $\bar{p}<\sup p$.

\begin{example} \rm Let
\begin{gather*}
p(t):= \begin{cases} {1\over2}-t& 0\le t\le 1;\\
t-{3\over 2}& 1<t\le{3\over 2};\\
0&  t>{3\over2},
\end{cases} \\
X(t):= \sin(t)\cos(1/2)-\cos(t)\sin(1/2)+3/2-t, \\
\begin{aligned}
Y(t)&:=2\sin(t)-4\sin(t)\cos(1/2)^2+\sin(t)\cos(1/2)-
\cos(t)\sin(1/2)\\
&\quad + 4\cos(t)\sin(1/2)\cos(1/2)-1/2+t,
\end{aligned} \\
\begin{aligned}
Z(t)&:=-2\sin(t)\cos(1/2)+2\sin(t)-4\sin(t)\cos(1/2)^2
+4\sin(t)\cos(1/2)^3\\
&\quad -4\cos(t)\sin(1/2)\cos(1/2)^2+
4\cos(t)\sin(1/2)\cos(1/2)+1,
\end{aligned}
\end{gather*}
 and $\bar{t}$ be the zero of $\dot{Z}$ in $[2.5,2.6]$.
Then
\[
x(t):=\begin{cases}
1&  0\le t\le {1\over2},\\
X(t)& {1\over2}< t\le 1,\\
Y(t)&  1<t<{3\over 2},\\
Z(t)& {3\over 2}<t\le\bar{t},\\
Z(\bar{t})&  t>\bar{t},
\end{cases}
\]
 solves \eqref{DE} for the above $p$, and
$Z(\bar{t})>{3\over4}$. Note that $\norm{p}_\infty={1\over 2}$,
whereas $\bar{p}=0$. In fact, every constant $t\mapsto \rho$ for
$\rho\in[-1,1]$ solves \eqref{DE} on $[{3\over2},\infty)$.
\end{example}

\section{Proof of Theorem \ref{MT}}
We proceed in several steps.

\noindent\textbf{Step 1.}  \emph{Let $x$ be a solution of \eqref{DE},
then $x$ cannot have a negative local maximum or a positive local
minimum on an interval which does not intersect dead zones.}

 In fact, if $a\in (0,\infty)$ and $x(a)$ is a
local minimum of $x$, then $x(a)^2=x(a)^2+\dot{x}(a)^2$. If $x(a)$
is positive, then $x(t)^2\ge x(a)^2$ for $t\in [a,a+\delta)$ and
some $\delta>0$. But, $t\mapsto x(t)^2+\dot{x}(t)^2$ is
nonincreasing by Lemma \ref{L1}(1); hence $x(t)=x(a)$ for
$t\in[a,a+\delta)$, i.e.~$x$ has a dead zone.

\noindent\textbf{Step 2.} {\it Let $x$ be a solution of \eqref{DE} and
$x_\infty:=\lim_{t\to\infty} x(t)$, which exists thanks to
Lemma \ref{L1}(4). If $\abs{x_\infty}< 1-\overline{p}$, then there
exists an $\underline{t}\in [0,\infty)$ with $x(t)=x_\infty$ for
$t\in [\underline{t},\infty)$.}

\noindent {\it Proof.} Select $\epsilon\in (0,1-\overline{p}
)$ with $\abs{x_\infty}<1-\overline{p}-4\epsilon$ and $\tilde{t}\in
(0,\infty)$ with $\abs{x(t)-x_\infty}<\epsilon$ and $\abs{\dot
x(t)}<\epsilon$ for $t\in[\tilde{t},\infty)$.

1-st case: $\dot x(\tilde{t})=0$. Noting that $\abs{p(t)-x(t)}\le
\overline{p}+\abs{x_\infty}+\epsilon\le \overline{p}+[1-\overline{p}
-4\epsilon]+\epsilon=1-3\epsilon$ for $t\in[\tilde{t},\infty)$, we
obtain $x(t)=x_\infty$ for $t\in[\tilde{t},\infty)$.

2-nd case: $\dot x(\tilde{t})>0$. Let $\underline{t}:=\sup \{t
\in[\tilde{t},\infty):\dot x(\tau)>0 \hbox{ for }\tau\in
[\tilde{t},t]\}$. Then $\ddot x(t)=p(t)-x(t)-1\le \overline{p}+
\abs{x_\infty}+\epsilon-1=\overline{p}+1-\overline{p}-4\epsilon
+\epsilon-1=-3\epsilon$ for $t\in[\tilde{t},\underline{t})$, hence
$\abs{\dot x(t)}<\epsilon$ for $t\in[\tilde{t},\infty)$ implies
$\underline{t}<\infty$. Since $\dot x(\underline{t})=0$, we arrive
at the first case with $\tilde{t}=\underline{t}$.

 Likewise, the last case $\dot x(\tilde{t})<0$ can be derived.

\noindent\textbf{Step 3.} {\it Let $x$ be a solution of \eqref{DE} with
$x_\infty:=\lim_{t\to\infty} x(t)\in
(1-\overline{p},1+\overline{p})$. Then $x$ has a dead zone in every
neighborhood of $\infty$.}

\noindent {\it Proof.} Otherwise, $\dot x$ possesses only
isolated zeroes in a neighborhood of $\infty$. By step 1, $x$ cannot
have a positive minimum, thus $x$ has to be monotone. Select
$\epsilon>0$ and $\underline{t}>0$ such that the following holds:

\begin{itemize}
\item $1-\overline{p}+\epsilon <x(t)<1+\overline{p}-\epsilon$ for
$t\ge\underline{t}$;

\item $\dot x$ has only isolated zeroes in
$[\underline{t},\infty)$;

\item $\abs{\dot x}<\epsilon$ on
$[\underline{t},\infty)$.
\end{itemize}

Assume that $x$ is monotone increasing on $[\underline{t},\infty)$.
Then $u(t)$ in (2) is equal to 1 almost everywhere on
$[\underline{t},\infty)$. Thus, $\ddot x(t)=p(t)-1-x(t)\le
\overline{p}-1-(1-\overline{p}+\epsilon)\le -\epsilon$ for a.e.
$t\in [\underline{t},\infty)$ which is a contradiction.

Likewise, $\ddot x(t)=p(t)-1-x(t)\ge
p(t)-1-(1+\overline{p}-\epsilon)\ge \epsilon$ for a.e. $t\in
[\underline{t},\infty)$ shows that $x$ cannot be decreasing near
$\infty$.

 Clearly, one obtains a corresponding assertion if
$\lim_{t\to\infty} x(t)\in (-1-\overline{p},\overline{p}-1)$.
\medskip

We remark that by Lemma \ref{L1}(1), $t\mapsto x(t)^2+\dot x(t)^2$
is nonincreasing, hence $x$ cannot increase under the assumptions of
step 3., when leaving a dead zone. Thus, $x$ is nonincreasing near
$\infty$.

\section{Some partial result for the case of multivalued systems}

In what follows, $\mathbf{a}\cdot\mathbf{b}$ denotes the Euclidian
scalar product of $\mathbf{a},\mathbf{b\in}\mathbf{R}^{N}$ and
$\|.\|$ the Euclidean norm.

A complete extension of Theorem \ref{MT} to systems of the form
\eqref{PN} appears to be quite difficult. Here we will merely
show that the solution
$$
\mathbf{x}(t):=(x_{1}(t),x_{2}(t),\dots,x_{N}(t))^{T}
$$
may develop a final dead zone $[\underline{t}, \infty)$.

We require stronger assumptions than those of the one-dimensional
case:
\begin{gather}
\mathbf{p}(t)^{T}\mathbf{\in}[-\frac{\mu_{\beta}}{2k}+\epsilon,\frac
{\mu_{\beta}}{2k}-\epsilon]^{N}\mbox{ for a.e. }t\geq T_{f},
\mbox{ for some }T_{f}\text{ and }\epsilon>0, \label{p vector}
\\
\Vert \mathbf{p}(t)-\mathbf{p}_{\infty}\Vert \to 0\quad \mbox{as}
t\to+\infty.\label{pinfinity}
\end{gather}


\begin{theorem}
We have:
\begin{itemize}
\item[(i)] Let $(u_{0},v_{0})\in \mathbb{R}^{2N}$, $p\in
L^{2}(0,\infty:\mathbb{R}^{N})$, and (\ref{p vector}) be satisfied. Then
problem (\ref{PN}) admits a unique weak solution $x\in
C^{1}([0,+\infty),\mathbb{R}^{N})$. If, moreover, (\ref{pinfinity}) holds,
then there exists a unique equilibrium state $x_{\infty}\in \mathbb{R}^{N}$,
i.e., $x_{\infty}$ satisfies
$Ax_{\infty}-p_{\infty}\in([-\frac{\mu_{\beta}}{2k},\frac{\mu_{\beta}}
{2k}]^{N})^{T}$, such that $\|\overset{\cdot}{\mathbf{x}
}(t)\|$ $+\| x(t)-x_{\infty}\|\to 0$
$\mbox{\emph{as} }t\to+\infty$.

\item[(ii)] Assume (\ref{p vector}) and (\ref{pinfinity}) hold. Let $x_{\infty}
$ be the associate equilibrium state and assume that
$$
|\Delta_{i}^{\ast}|<1\text{ \textit{where} }\Delta_{i}^{\ast}:=(\mathbf{Ax}%
_{\infty})_{i}-p_{i.\infty},\quad\emph{for\ some\ }i\in{1,\dots,N}.
$$
 Then there exists $T_{i}\geq 0$ such that $\dot{x}_{i}(t)=0$ for
 all $t\geq T_{i}$.
\end{itemize}
\end{theorem}

\begin{proof} We shall adapt in our presentation some arguments
of \cite{DM}. To reformulate (\ref{PN}) in the framework of
nonlinear semi-group operators theory, we introduce the \emph{phase
space}
$\mathbf{H}=(\mathbb{R}^{N},\langle,\rangle_{\mathbf{A}})
\times(\mathbb{R}^{N},\cdot)$,
with $\langle \mathbf{a,b}\rangle_{\mathbf{A}}=\mathbf{Aa\cdot b}$, where
$\mathbf{A}$ is the symmetric positive definite matrix of
$\mathbb{R}^{N\times N}$ given by
\[
\mathbf{A=}
\begin{pmatrix}
2 & -1 & 0 & ... & 0\\
-1 & 2 & -1 & 0 & ...\\
0 & -1 & 2 & -1 & 0\\
... & 0 & -1 & 2 & -1\\
0 & ... & 0 & -1 & 2
\end{pmatrix}
\]
We also introduce
$\mathbf{B:}\mathbf{R}^{N}\to\mathcal{P}(\mathbb{R}^{N})$ as the
(multivalued) maximal monotone operator
given by
$\mathbf{B}(y_{1},\dots,y_{N})=(\beta(y_{1}),\dots,\beta(y_{N}))^{T}$
 where $\beta(s)=Sgn(s)$. Finally, we define the operator
$\mathbf{L}$ in $\mathbf{H}$ by
\begin{equation}
\mathbf{L(x,y)}=\{\mathbf{-y}\}\times\{\frac{k}{m}\mathbf{Ax}+\frac{\mu
_{\beta}}{m}\mathbf{B}(\mathbf{y})\}\text{ for }\mathbf{(x,y)\in H.}%
\end{equation}
It is easy to prove that $\mathbf{L}$ is maximal monotone in $H$ and
so, by using results from the theory of maximal monotone operators
(see \cite{Brezis-maximaux}) we get the existence and uniqueness of
a solution of (\ref{PN}). Multiplying the equation by
$\overset{\cdot}{\mathbf{x}}(t)$ and integrating in time we get the
\emph{energy relation}
\begin{equation}
E(t)+\int_{0}^{t}[\sum_{i=1}^{N}\frac{\mu_{\beta}}{m}|\dot{x}_{i}%
(s)|-p_{i}(s)\dot{x}_{i}(s)]ds=E(0)\label{Energia}%
\end{equation}
where
\begin{equation}
E(t)=\frac{1}{2}\|\overset{\cdot}{\mathbf{x}}(t)\|^{2}+\int
_{0}^{t}\frac{k}{2m}\mathbf{Ax}(s)\cdot\mathbf{x}(s)ds.
\end{equation}
By (\ref{Energia}) and the assumptions on $\mathbf{p}(t)$, the
trajectory $(\mathbf{x}(t),\overset{\cdot}{\mathbf{x}}(t))_{t\geq0}$
is compact in
$\mathbf{H}$, so, we can find $\alpha>0$ such that $\mu_{\beta}|\dot{x}%
_{i}(t)|-p_{i}(s)\dot{x}_{i}(t)\geq\alpha|\dot x_{i}(t)|$ for
$i=1,\dots,N$
and all $t\geq0$. By (\ref{Energia}), we conclude that $\dot{\mathbf{x}%
}\in L^{1}({\mathbb{R}}_{+})$ which leads to the existence of the
limit $\mathbf{x}_{\infty}:=\lim_{t\to+\infty}\mathbf{x}(t)$
and to $\lim_{t\to+\infty}\dot{\mathbf{x}}(t)=0$. The
uniqueness of $\mathbf{x}_{\infty}$ is deduced from the strict
monotonicity of the operator
$\widetilde{\mathbf{L}}\mathbf{(x,y)}=\{\mathbf{-y}\}\times\{\frac{k}%
{m}\mathbf{Ax}\}$ for $\mathbf{(x,y)\in H}$.

To prove (ii) we recall that, since
$\mathbf{x}_{\infty}$ is an stationary point, we have
$(\Delta_{i}^{\ast})_{i=1}^{N}\in\lbrack-1,1]^{N}$. Now,\textit{
}let $0<\delta<<1$ be fixed. By (i) we can find $t_{0}\geq0$ such
that
\begin{equation}
|\Delta_{i}(t)|\leq(1-2\delta)\quad\forall t\geq t_{0},\label{decay}%
\end{equation}
If $\dot{x}_{i}(t_{0})=0$, we conclude that $x_{i}(t)\equiv x_{i}%
(t_{0})=x_{\infty i}$ for all $t\geq t_{0}$ since $\Delta_{i}(t)\in
\lbrack-1,1]$ for all $t\geq t_{0}$. If not, let $T=\sup\{s\geq t_{0}%
,\;|\dot{x}_{i}(t)|>0\;\forall t\in [t_{0},s)\}$. Multiplying the
i-component of (\ref{PN}) by $\dot{x}_{i}(t)$ and using
(\ref{pinfinity}) we obtain
\begin{equation}
\frac{1}{2}\frac{d}{dt}(|\dot{x}_{i}(t)|^{2})+\delta|\dot{x}_{i}%
(t)|\leq0,\quad \text{for a.e. }t\in [t_{0},T).\label{desig-energia}
\end{equation}
Dividing (\ref{desig-energia}) by $|\dot{x}_{i}(t)|$ we get
\begin{equation}
\frac{d}{dt}(|\dot{x}_{i}(t)|)+\delta\leq0\quad \text{ for a.e. }t\in [
t_{0},T).
\end{equation}
Integrating, we see that
$\dot{x}_{i}(t_{0}+\frac{|\dot{x}_{i}(t_{0})|}{\delta})=0.$ Thus $T<+\infty$
 and we conclude, as before, that
$x_{i}(t)\equiv x_{i}(T)=x_{\infty i}$ for any $t\geq T$.
\end{proof}


We remark that the behavior exhibited in the above result
is different from the case in which the amplitude of
$\mathbf{p(}t\mathbf{)}$ becomes large. In that case the dynamics
may generate a wide range of events leading to chaos (see
\cite{Carlson}).


\subsection*{Acknowledgements}
 The second author (G. Hetzer) wants to thank
the department of applied mathematics of Complutense University for
supporting his stay in June 2006.

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\end{document}